\documentclass[11pt, a4paper]{article} \title{\vspace{-2cm}\bf Supplementary Information:\\ {\Large Quantum Zeno-like Paradox for Position Measurements}\vspace{-0.25cm}} \author{Xabier Oianguren-Asua\footnote{E-mail: xabier.oianguren-asua@math.uni-tuebingen.de} ~~and Roderich Tumulka\vspace{0.2cm}\\ {\small \em Mathematics Institute, Eberhard Karls University T\"ubingen,}\\ {\small \em Auf der Morgenstelle 10, 72076 T\"ubingen, Germany. }} \date{\vspace{-0.5cm}} \usepackage[utf8]{inputenc} \usepackage[top=2.3cm, bottom=2.3cm, left=2.3cm, right=2.3cm]{geometry} \usepackage{amsmath} \usepackage{amsthm} %theorems \usepackage{amssymb}\usepackage{mathrsfs} \DeclareMathAlphabet{\mathpzc}{OT1}{pzc}{m}{it} \usepackage{enumerate} \usepackage{dsfont} % indicator function's 1 \usepackage{physics} \usepackage[backend=biber, style=ieee]{biblatex} \addbibresource{Bibliography.bib} \renewcommand*{\thefootnote}{[\arabic{footnote}]} \setlength{\footnotesep}{\baselineskip} \newcommand{\atright}[1]{% \unskip % remove a possible space in front \hfil % glue for centering \makebox[0pt][r]{#1}% the tag at the right \hfilneg % to countermand the \rightskip \hspace{0pt}% \par removes a glob of glue \par % ensure the line is ended } \usepackage[% framemethod=tikz, linewidth=0.01cm, middlelinecolor= gray, middlelinewidth=0.02cm, roundcorner=1cm, topline = false, rightline = false, bottomline = false, skipabove=0pt, skipbelow=0pt, usetwoside=false, leftmargin=-0.6cm, innerleftmargin=0.6cm, rightmargin=-0.3cm, linecolor=gray,% backgroundcolor=none ]{mdframed} \newmdenv[tikzsetting={draw=gray,fill=red,fill opacity=0.5}]{myenvironment} \theoremstyle{definition} \newtheorem{definition}{Definition} \newtheorem{lemma}{Lemma} \newtheorem{theorem}{Theorem} \newenvironment{Lemma}[1] {\renewcommand\thelemma{#1}\lemma} {\endlemma} \newenvironment{Definition}[1] {\renewcommand\thedefinition{#1}\definition} {\enddefinition} \newenvironment{cproof}[1][\unskip]{\begin{mdframed} \noindent {\em Proof #1: }\renewcommand{\thempfootnote}{[\alph{mpfootnote}]}}{\vspace{-0.3cm}\flushright{\qedsymbol}\end{mdframed} } \renewcommand{\thefootnote}{(\alph{footnote})} \newenvironment{gproof}[1][\unskip]{\begin{myenvironment} \noindent {\em Proof #1: }}{\vspace{-0.3cm}\flushright{\qedsymbol}\end{myenvironment} } \newcommand{\Lozenge}{\vspace{-0.7cm}\flushright{{\tiny $\blacklozenge$}}} \usepackage{xcolor} \definecolor{tuered}{rgb}{0.627, 0.082, 0.227} \usepackage{hyperref} \hypersetup{ colorlinks=true, linkcolor=tuered, filecolor=tuered, urlcolor=tuered, citecolor=tuered } \urlstyle{same} \usepackage[skip=0.2cm, indent=0.3cm]{parskip} \renewcommand{\baselinestretch}{1.15} \newcommand{\RT}[1]{{\color{blue}{#1}}} % indicating changes \newcommand{\XOA}[1]{{\color{RoyalPurple}{#1}}} % indicating changes \usepackage{lineno} \setlength{\linenumbersep}{1cm} \begin{document} \maketitle \linenumbers \renewcommand{\theequation}{S\arabic{equation}} \section*{Mathematical Proofs} \nocite{Misra77, tumulka, Bur, rigged_original, rigged_Gadella, Ber, Wer} Since Theorem~1 is a special case of Theorem~2, it suffices to prove the latter. It is well known that because $Q$ has finite volume, $L^\infty(Q) \subseteq L^2(Q)\subseteq L^1(Q)$.\vspace{0.3cm} \begin{Definition}{S1} Given $f\in L^1(Q)$ and a grid scheme $\{B_j^{(n)}\}_{j=1}^{N_n}$ for $Q$, we define the ``discretization of $f$ by the $n$-th grid" to be the function $f_n:Q\rightarrow\mathbb{C}$, \begin{equation} f_n(x):=\sum_{j=1}^{N_n}\underbrace{\frac{1}{|B_j^{(n)}|}\int_{y\in B_j^{(n)}} f(y) \,dy}_{\text{average of $f$ in $B_j^{(n)}$}}\;\mathds{1}_{B_j^{(n)}}(x)\quad \quad \text{ for } x\in Q.\vspace{-0.2cm} \end{equation}(Informally, it is the ``bar chart" taking at each point of $B_j^{(n)}$ the average of $f$ in $B_j^{(n)}$ as value.) \end{Definition}\vspace{0.6cm} \begin{Lemma}{S1}\label{leb.diff.lemma} {\it Let there be an arbitrary $f\in L^\infty(Q)$ and let $\{B_j^{(n)}\}_{j=1}^{N_n}$ be an arbitrary grid scheme for $Q$. Then, $f_n\in L^{{\infty}}(Q)$ and $\norm{f_n-f}_{L^2}\xrightarrow[n\rightarrow\infty]{}0$.} \end{Lemma} \begin{proof} First, note that for each $x\in Q$ and $n\in \mathbb{N}$, because $\{B_j^{(n)}\}_{j=1}^{N_n}$ is a partition of $Q$, there exists a unique $j_{x}^n\in \{1,...,N_n\}$ such that $x\in B_{j_x^n}^{(n)}$. Next, for each $x\in Q$,\vspace{-0.3cm} \begin{align} | f_n(x)|&=\Bigg|\sum_{j=1}^{N_n}\frac{\mathds{1}_{B_j^{(n)}}(x)}{|B_j^{(n)}|}\int_{y\in B_j^{(n)}} f(y) \, dy\Bigg| \end{align} \begin{align} &\leq \sum_{j=1}^{N_n}\frac{\mathds{1}_{B_j^{(n)}}(x)}{|B_j^{(n)}|}\int_{y\in B_j^{(n)}} |f(y)| \, dy\\ &\stackrel{\ref{only.one.survives}}{=}\frac{1}{|B_{j_x^n}^{(n)}|}\int_{y\in B_{j_x^n}^{(n)}} |f(y)| \, dy\\ &\leq \norm{f}_{L ^\infty}, \end{align} \stepcounter{footnote} \footnotetext{\label{only.one.survives}By definition, $\mathds{1}_{B_j^{(n)}}(x)=0$ for all $j\in \{1,...,N_n\}$ except for $j=j_x^n$, where $\mathds{1}_{B_{j_x^n}^{(n)}}(x)=1$. } and hence $\norm{f_n}_{L^\infty}\leq \norm{f}_{L^\infty}$, such that $f_n\in L^\infty(Q)$, and a fortiori $f_n\in L^2(Q)$. Now recall that for each $x\in Q$, $B_{j_x^n}^{(n)}=\prod_{k=1}^nI_{j_x^n,k}^{(n)}$ for some intervals $I_{j_x^n,k}^{(n)}\subseteq[0,1)$ of length in $[\frac{1}{Cn}, \,\frac{1}{n}]$. This implies that the longest diagonal of $B_{j_x^n}^{(n)}$ has at most length $\frac{\sqrt{d} }{n}$ and thus that $B_{j_x^n}^{(n)}$ is contained in the open ball of radius $R_n:=\frac{2\sqrt{d} }{n}$ around $x$, which we denote by $V_{R_n}(x)$. Moreover, for the $n$-independent number $K:=\frac{1}{C^d(2\sqrt{d} )^d|V_{1}(0)|}$, \newpage \begin{equation} |B_{j_x^n}^{(n)}|\stackrel{\footnote{By definition, the edges of $B_{j_x^n}^{(n)}$ have a length no smaller than $1/(Cn)$, so the volume of $B_{j_x^n}^{(n)}$ is at least $\frac{1}{(Cn)^d}$. }}{\geq} \frac{1}{(Cn)^d}= K\Big(\frac{2\sqrt{d} }{n}\Big)^d|V_{1}(0)|= K (R_n)^d|V_1(0)|=K|V_{R_n}(x)|. \end{equation} This condition, together with the fact that $f$ is integrable\footnote{By assumption, $f\in L^\infty(Q)$, which implies that $f\in L^1(Q)$.} ---which by Theorem 1.6.19 in Ref.~\cite{tao} implies that almost every (a.$\;$e.) $x\in Q$ is a Lebesgue point--- allow us to apply Ex.$\;$1.6.15 in Ref.~\cite{tao}, which is a corollary of the Lebesgue differentiation theorem. This corollary proves that for a.$\;$e.$\;x\in Q$, \begin{equation}\label{Lebsegue.diff.thm} \lim_{n\rightarrow\infty}\Bigg(\frac{1}{|B_{j^n_x}^{(n)}|}\int_{y\in B_{j_x^{n}}^{(n)}}f(y) \, dy\Bigg)=f(x). \end{equation} Now, note the following.\begin{enumerate} \item[(i)] Point-wise, for a.$\;$e.$\;x\in Q$,\vspace{-0.1cm} \begin{align} |f_n(x)-f(x)|&=\Bigg|\sum_{j=1}^{N_n}\frac{\mathds{1}_{B_j^{(n)}}(x)}{|B_j^{(n)}|}\int_{y\in B_j^{(n)}}f(y)\, dy -f(x)\Bigg|\\ &\stackrel{\ref{only.one.survives}}{=}\Bigg|\frac{1}{|B_{j_x^n}^{(n)}|}\int_{y\in B_{j_x^n}^{(n)}}f(y)\, dy -f(x)\Bigg|\xrightarrow[n\rightarrow\infty]{\eqref{Lebsegue.diff.thm}}0. \end{align} \item[(ii)] The constant (and hence $L^\infty)$ function $g(x)\equiv 2\norm{f}_{L^\infty}$ satisfies that for a.$\;$e.$\;x\in Q$, \begin{equation} |f_n(x)-f(x)|\leq \frac{1}{|B_{j_x^n}^{(n)}|}\int_{y\in B_{j_x^n}^{(n)}}\Big|f(y)\Big| \, dy +\Big|f(x)\Big|\leq 2\norm{f}_{L^\infty}\:. \end{equation} \end{enumerate} By virtue of (i) and (ii), $g$ is a dominating function in $L^2(Q)$ that grants the application of the dominated convergence theorem in the last step of the following chain of equations: \begin{equation} \lim_{n\rightarrow\infty}\norm{f_n-f}_{L^2}^{2}=\lim_{n\rightarrow\infty}\int_{x\in Q}|f_n(x)-f(x)|^{2} \, dx=0.\vspace{-0.3cm} \end{equation} \end{proof} We are now ready to prove the result for $Q$ in the special case that $\hat\rho=|\psi\rangle\langle\psi|$ is pure and both $\psi$ and $\phi$ are (essentially) bounded functions rather than general $L^2$ functions.\vspace{0.3cm} \begin{Lemma}{S2}\label{result.for.Linfty} {\it Given arbitrary $\psi,\phi\in L ^\infty(Q)$ and an arbitrary grid scheme $\{B_j^{(n)}\}_{j=1}^{N_n}$ for $Q$:} \vspace{-0.1cm} \begin{equation} \lim_{n\rightarrow\infty}\sum_{j=1}^{N_n}\Big|\langle \phi| \hat{P}_j^{(n)}\psi\rangle\Big|^2=0. \end{equation} \end{Lemma} \begin{proof} First, define $f(x):=\overline{\phi(x)}\psi(x)$ for $x\in Q$. Then $f\in L ^\infty(Q)$ because for a.$\,$e.$\;x\in Q$, $|f(x)|=|\phi(x)| \, |\psi(x)|\leq \|\phi\|_{L^\infty} \|\psi\|_{L^\infty}$. %\frac{1}{2}|\phi(x)|^2+\frac{1}{2}|\psi(x)|^2\leq \frac{1}{2}(\norm{\phi}_{L ^\infty}^2+\norm{\psi}_{L^\infty}^2). Hence, $f\in L^2(Q)$ and moreover, we can apply Lemma \ref{leb.diff.lemma} to $f$ to get that $\norm{f_n-f}_{L^2}\xrightarrow[n\rightarrow \infty]{}0$. This implies by the reverse triangle inequality that $\norm{f_n}_{L^2}\rightarrow\norm{f}_{L^2}$ and thus $\norm{f_n}_{L^2}^2\rightarrow\norm{f}_{L^2}^2.$ Now, \begin{align} \norm{f_n}_{L^2}^2&=\int_{x\in Q}\Bigg| \sum_{j=1}^{N_n}\frac{\mathds{1}_{B_j^{(n)}}(x)}{|B_j^{(n)}|}\underbrace{\int_{y\in B_j^{(n)}} f(y) \, dy}_{\int_{y\in B^{(n)}_j}\overline{\phi(y)} \, \psi(y) \, dy}\;\Bigg|^2dx\\ &=\int_{x\in Q}\Bigg| \sum_{j=1}^{N_n}\frac{\mathds{1}_{B_j^{(n)}}(x)}{|B_j^{(n)}|}\langle \phi| \hat{P}_j^{(n)}\psi\rangle\;\Bigg|^2dx\\ \intertext{[use that $\{\mathds{1}_{B_j^{(n)}}\}_{j=1}^{N_n}$ have disjoint supports]} &= \sum_{j=1}^{N_n}\int_{x\in Q}\Bigg| \frac{\mathds{1}_{B_j^{(n)}}(x)}{|B_j^{(n)}|}\langle \phi| \hat{P}_j^{(n)}\psi\rangle\;\Bigg|^2dx \label{stepstar}\\ &=\sum_{j=1}^{N_n}\frac{|B_j^{(n)}|}{|B_j^{(n)}|^2} \bigl|\langle\phi| \hat{P}^{(n)}_j\psi\rangle \bigr|^2\\ &=\sum_{j=1}^{N_n}\frac{1}{|B_j^{(n)}|} \bigl|\langle\phi| \hat{P}^{(n)}_j\psi\rangle \bigr|^2. \end{align} But then $\lim_{n\rightarrow\infty}\norm{f_n}_{L^2}^2=\norm{f}^2_{L^2}$ means that the sequence $\Big(\sum_{j=1}^{N_n} \frac{1}{|B_j^{(n)}|}\Big|\langle \phi| \hat{P}_j^{(n)}\psi\rangle\Big|^2\Big)_{n\in \mathbb{N}}$ converges to the finite number $\norm{f}_{L^2}^2$. Using this result to obtain Eq.~\eqref{stepstarstar} below, together with the fact that $(1/n^d)_{n\in \mathbb{N}}$ converges to 0 and the continuity of the product of complex numbers, \begin{align} \sum_{j=1}^{N_n}\Big|\langle \phi| \hat{P}_j^{(n)}\psi\rangle\Big|^2 &=\sum_{j=1}^{N_n}\frac{|B_j^{(n)}|}{|B_j^{(n)}|}\Big|\langle \phi| \hat{P}_j^{(n)}\psi\rangle\Big|^2\\ &\stackrel{(12)}{\leq} \frac{1}{n^d}\ \sum_{j=1}^{N_n} \frac{1}{|B_j^{(n)}|}\Big|\langle \phi| \hat{P}_j^{(n)}\psi\rangle\Big|^2\\ &\xrightarrow[n\rightarrow\infty]{}0\cdot\norm{f}_{L^2}^2=0. \label{stepstarstar} \end{align}\vspace{-0.3cm} \end{proof} With that, we can prove the spatial quantum Zeno effect for any pure state $\hat{\rho}=|\psi\rangle\langle \psi|$ on $L^2(Q)$.\vspace{0.3cm} \begin{Lemma}{S3}\label{main.result.in.Q.for.pure.states} {\it Given arbitrary $\psi,\phi\in L ^2(Q)$ and an arbitrary grid scheme $\{B_j^{(n)}\}_{j=1}^{N_n}$ for $Q$},\vspace{-0.1cm} \begin{equation} \lim_{n\rightarrow\infty}\sum_{j=1}^{N_n}\Big|\langle \phi| \hat{P}_j^{(n)}\psi\rangle\Big|^2=0.\vspace{-0.1cm} \end{equation} \end{Lemma} \begin{proof} It is well known that the smooth compactly supported functions $\mathcal{C}^\infty_0(Q)$ are dense in $L^2(Q)$ (see Prop.$\;$8.17 in Ref.~\cite{folland} for the case of $\mathbb{R}^d$ ---with the right identifications, one can restrict this result to $Q$). In particular, all functions on $\mathcal{C}_0^\infty(Q)$ are continuous maps with compact support and thus, they are bounded functions. As such, $\mathcal{C}_0^\infty(Q)\subseteq L^\infty(Q)$, proving that $L^\infty(Q)$ is also dense in $L^2(Q).$ By this denseness, for an arbitrary $\varepsilon>0$, one can successively find $\psi_\varepsilon,\phi_\varepsilon \in L^\infty(Q)$ such that $\norm{\psi-\psi_\varepsilon}^2\leq \frac{\varepsilon}{6\norm{\phi}^2}$ and $\norm{\phi-\phi_\varepsilon}^2\leq \frac{\varepsilon}{12\norm{\psi_\varepsilon}^2}.$ In particular, by Lemma \ref{result.for.Linfty}, $\sum_{j=1}^{N_n} \frac{1}{|B_j^{(n)}|}\Big|\langle \phi_\varepsilon| \hat{P}_j^{(n)}\psi_\varepsilon\rangle\Big|^2$ $\xrightarrow[n\rightarrow \infty]{}0$, so there is $n_\varepsilon\in \mathbb{N}$ such that $\sum_{j=1}^{N_n} \frac{1}{|B_j^{(n)}|}\Big|\langle \phi_\varepsilon| \hat{P}_j^{(n)}\psi_\varepsilon \rangle\Big|^2\leq \frac{\varepsilon}{12 }$ for all $n\geq n_\varepsilon.$ Hence, for all $n\geq n_\varepsilon$, \begin{align} \sum_{j=1}^{N_n}\Big|\langle & \phi| \hat{P}^{(n)}_j\psi\rangle\Big|^2\ \stackrel{\ref{triangl.ineq.sqr}}{\leq}\ \sum_{j=1}^{N_n}2\Big(\Big|\langle \phi| \hat{P}^{(n)}_j(\psi-\psi_\varepsilon)\rangle\Big|^2+\Big|\langle \phi| \hat{P}^{(n)}_j\psi_\varepsilon\rangle\Big|^2\Big)\\ &\stackrel{\ref{triangl.ineq.sqr}}{\leq} \sum_{j=1}^{N_n}2\Big(\Big|\langle \hat{P} ^{(n)}_j\phi| \psi-\psi_\varepsilon\rangle\Big|^2+2\Big|\langle \phi-\phi_\varepsilon| \hat{P}^{(n)}_j\psi_\varepsilon\rangle\Big|^2+2\Big|\langle \phi_\varepsilon| \hat{P}^{(n)}_j\psi_\varepsilon\rangle\Big|^2\Big)\\ &\leq 2\underbrace{\sum_{j=1}^{N_n}\norm{\hat{P}^{(n)}_j\phi}^2}_{=\norm{\phi}^2\text{ by \ref{id.resolution}}}\norm{\psi-\psi_\varepsilon}^2+4\norm{\phi-\phi_\varepsilon}^2\underbrace{\sum_{j=1}^{N_n} \norm{\hat{P}^{(n)}_j\psi_\varepsilon}^2}_{=\norm{\psi_\varepsilon}^2\text{ by \ref{id.resolution}}}+4\sum_{j=1}^{N_n}\frac{|B_j^{(n)}|}{|B_j^{(n)}|}\Big|\langle \phi_\varepsilon| \hat{P}^{(n)}_j\psi_\varepsilon\rangle\Big|^2\\ \intertext{[now use the definition of $\psi_\varepsilon,\phi_\varepsilon$ and Eq.~(12)]} &\leq\ \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+ 4\;\frac{1}{n^d}\sum_{j=1}^{N_n}\frac{1}{|B_j^{(n)}|}\Big|\langle \phi_\varepsilon| \hat{P}^{(n)}_j\psi_\varepsilon\rangle\Big|^2\ \stackrel{(n\geq n_\varepsilon)}{\leq}\ \frac{2\varepsilon}{3}+\frac{1}{n^d}\frac{\varepsilon}{3}\ \leq \ \varepsilon. \end{align} \stepcounter{footnote} \footnotetext{\label{triangl.ineq.sqr} Successively, we add and subtract $\langle \phi| \hat{P}^{(n)}\psi_\varepsilon\rangle$ and then $\langle \phi_\varepsilon| \hat{P}^{(n)}\psi_\varepsilon\rangle$, using after each such operation that for all $a,b\in\mathbb{C}$, $|a-b|^2=|a|^2+|b|^2-2\mathrm{Re}\{\overline{a}b\}\leq |a|^2+|b|^2+2|a||b|\leq 2|a|^2+2|b|^2$. } \stepcounter{footnote} \footnotetext{\label{id.resolution}Let $\eta\in L^2(Q)$ be arbitrary. Then, using that $\hat{P}_j^{(n)}$ are self-adjoint projectors and the obvious property $\hat{I}=\sum_{j=1}^{N_n}\hat{P}_j^{(n)}$, we get that $\sum_{j=1}^{N_n}\norm{\hat{P}_j^{(n)}\eta}^2=\langle \eta| \sum_{j=1}^{N_n}\hat{P}_j^{(n)}\eta\rangle=\langle \eta|\eta\rangle=\norm{\eta}^2.$} But by definition, this is to say that $\displaystyle \sum_{j=1}^{N_n}\Big|\langle \phi| \hat{P} ^{(n)}_j\psi\rangle\Big|^2\xrightarrow[n\rightarrow\infty]{}0$.\vspace{-0.5cm} \end{proof} \vspace{0.3cm} The generalization of Lemma \ref{main.result.in.Q.for.pure.states} to translated and scaled cubes (namely, to arbitrary rectangles in $\mathbb{R}^d$) is straightforward. Next, we are going to split $\mathbb{R}^d$ into countably many cubes and equip each of them with a grid scheme, making together a family of ``position measurement grids" for the whole $\mathbb{R}^d$. \vspace{0.3cm} \begin{Definition}{S2}\label{def.partition.Rd} Given a translated unit cube $\widetilde{Q}:=\prod_{k=1}^d[a_k,\ a_k+1)$ for some $a_k\in\mathbb{R}$, we define a {\em grid scheme for $\widetilde{Q}$} to be the set of translations $\widetilde{V}_{j}^{(n)}:={V}_{j}^{(n)}+(a_1,...,a_d)$ of an arbitrary grid scheme for $[0,1)^d$, $\{{V}_{j}^{(n)}\}_{j=1}^{N_n}$. \noindent Then, we define a {\em grid scheme for $\mathbb{R}^d$} to be a partition of $\mathbb{R}^d$ into translated unit cubes $\{Q_\ell\}_{\ell\in \mathbb{N}}$ and a choice of {\em grid scheme} per $Q_\ell$, say, $\{\widetilde{V}_{\ell,j}^{(n)}\}_{\ell\in\mathbb{N},\ j\in\{1,...,N_\ell^{(n)}\}}$ $(n\in\mathbb{N})$. Since for each $n\in\mathbb{N}$, $\{V_{\ell,j}^{(n)}\}_{\ell\in\mathbb{N},\ j\in\{1,...,N_\ell^{(n)}\}}$ is a countable set, we will write $\{V_{\ell,j}^{(n)}\}_{\ell\in\mathbb{N},\ j\in\{1,...,N_\ell^{(n)}\}}=\{B_j^{(n)}\}_{j\in\mathbb{N}}$ and define $J_\ell^{(n)}:=\{j\in \mathbb{N}\ |\ B_j^{(n)}\subseteq Q_\ell\}$ (such that $|J_\ell^{(n)}|=N_\ell^{(n)}$).\vspace{0.2cm}\Lozenge \end{Definition} For each fixed $n\in\mathbb{N}$, $\{B_j^{(n)}\}_{j\in\mathbb{N}}$ is trivially a partition of $\mathbb{R}^d$ into rectangles of edges in the order of $1/n$. The only restriction we put by the inductive definition is that the interior of no $B_j^{(n)}$ is allowed to intersect the boundary of a cube $Q_\ell$. Although this made the last definition a bit cumbersome, we imposed it to make the following statements and their proofs notationally simpler. \vspace{0.3cm} \begin{Lemma}{S4}\label{general.result.in.Rd.for.pure} {\em Given arbitrary $\psi,\phi\in L^2(\mathbb{R}^d)$ and an arbitrary grid scheme $\{B_{j}^{(n)}\}_{j\in \mathbb{N}}$ for $\mathbb{R}^d$}: \begin{equation} \lim_{n\rightarrow\infty}\sum_{j=1}^\infty\Big|\langle \phi|\hat{P}_j^{(n)}\psi\rangle \Big|^2=0. \end{equation} \end{Lemma} \begin{proof} First, note the following observation. Given a measurable infinite set $A\subseteq\mathbb{R}^d$, a partition $\{A_M\}_{M\in\mathbb{N}}$ for $A$ and a non-negative integrable function $f:\mathbb{R}^d\rightarrow[0,+\infty]$, \begin{enumerate} \item[(i)] for all $x\in A$, $\sum_{j=1}^M\mathds{1}_{A_j}(x) f(x)$ equals $\mathds{1}_A(x) f(x)$ from some $M$ onward,\footnote{Namely, for all $M$ if $x\not\in A$ and otherwise, after the $M$ such that $x\in A_M$.} \item[(ii)] $\sum_{j=1}^M\mathds{1}_{A_j} f$ is a monotonously increasing sequence of measurable functions. \end{enumerate} Consequently, the monotone convergence theorem implies that: \begin{equation}\label{claim} \lim_{M\rightarrow\infty} \sum_{j=1}^M \Big\|\mathds{1}_{A_n} f\Big\|_{L^1(\mathbb{R^d})}=\lim_{M\rightarrow\infty}\int_{x\in \mathbb{R}^d}\sum_{j=1}^M\mathds{1}_{A_n}(x)f(x)dx=\int_{x\in\mathbb{R}^d}\mathds{1}_A(x)f(x)dx=\Big\|\mathds{1}_A f\Big\|_{L^1(\mathbb{R}^d)}. \end{equation} This proves the last step of the following. \begin{align} \sum_{j=1}^M|\langle \phi| \hat{P}_j^{(n)}\psi\rangle |^2&\leq \|\phi\|^2\sum_{j=1}^M\|\hat{P}_j^{(n)}\psi\|^2\quad\\ &=\|\phi\|^2\sum_{j=1}^M\Big\|\mathds{1}_{B_j^{(n)}}|\psi|^2\Big\|_{L^1}\ \ \xrightarrow[M\rightarrow\infty]{\eqref{claim}}\ \ \|\phi\|^2\|\psi\|^2. \end{align} Hence, for each $n\in\mathbb{N}$, the symbol $\sum_{j=1}^\infty\Big|\langle \phi|\hat{P}_j^{(n)}\psi\rangle \Big|^2$ in the Lemma's statement is finite and well defined. In particular, it is an absolutely convergent series, which implies that one can add its terms in finite ``chunks" before adding them together and the result will still be the same. That is, using the notation of Definition \ref{def.partition.Rd}, \begin{equation}\label{in.chunks} \sum_{j=1}^\infty|\langle \phi| \hat{P}_j^{(n)}\psi\rangle |^2=\sum_{\ell=1}^\infty\sum_{j\in J_\ell^{(n)}} |\langle \phi| \hat{P}_j^{(n)}\psi\rangle |^2. \end{equation} Now, define $h_n(\ell):=\sum_{j\in J_\ell^{(n)}} |\langle \phi| \hat{P}_j^{(n)}\psi\rangle |^2$ and note the following. \begin{enumerate}[(i)] \item Point-wise, for each fixed $\ell\in \mathbb{N}$, \begin{equation} \lim_{n\rightarrow\infty}h_n(\ell)\ =\ \lim_{n\rightarrow\infty}\sum_{j\in J_\ell^{(n)}} |\langle \phi| \hat{P}_j^{(n)}\psi\rangle |^2\ \stackrel{\text{(Lemma \ref{main.result.in.Q.for.pure.states})}}{=}\ 0. \end{equation} \item For each $\ell\in\mathbb{N}$ and independently of $n\in\mathbb{N}$, \begin{equation} |h_n(\ell)|\ \leq\ \|\phi\|^2\sum_{j\in J_\ell^{(n)}}\|\hat{P}_j^{(n)}\psi\|^2\ \stackrel{\eqref{claim}}{=}\ \|\phi\|^2\Big\|\mathds{1}_{Q_\ell}|\psi|^2\Big\|_{L^1(\mathbb{R}^d)}\ =:g(\ell). \end{equation} In particular, denoting by $d\nu$ the counting measure of $\mathbb{N}$, \begin{equation} \|g\|_{L^1(\mathbb{N},d\nu)}\ =\ \|\phi\|^2\lim_{M\rightarrow\infty}\sum_{\ell=1}^M\Big\|\mathds{1}_{Q_\ell}|\psi|^2\Big\|_{L^1(\mathbb{R}^d)}\ \stackrel{\eqref{claim}}{=}\ \|\phi\|^2\|\psi\|^2\ <+\infty. \end{equation} Hence, $g$ is a dominating function in $ L^1(\mathbb{N},d\nu)$ for the sequence of functions $h_n:\mathbb{N}\rightarrow\mathbb{R}$. \end{enumerate} By the dominated convergence theorem, (i) and (ii) imply that \begin{equation}\label{dominated.conv.Rd} \lim_{n\rightarrow\infty}\int_{\ell\in \mathbb{N}}h_n(\ell)d\nu=0. \end{equation} Putting it all together, \begin{equation} \lim_{n\rightarrow\infty} \sum_{j=1}^\infty|\langle \phi| \hat{P}_j^{(n)}\psi\rangle |^2\ \stackrel{\eqref{in.chunks}}{=}\ \lim_{n\rightarrow\infty}\sum_{\ell=1}^\infty\sum_{j\in J_\ell^{(n)}} |\langle \phi| \hat{P}_j^{(n)}\psi\rangle |^2\ \stackrel{\text{(by def.)}}{=}\ \lim_{n\rightarrow\infty}\int_{\ell\in \mathbb{N}}h_n(\ell)d\nu\ \stackrel{\eqref{dominated.conv.Rd}}{=}\ 0.\vspace{-0.5cm} \end{equation} \end{proof} \vspace{0.3cm} Finally, let us lift these results to the case of arbitrary density matrices.\vspace{0.3cm} \noindent {\bf Theorem 2.} {\em For any positive trace-class operator $\hat{\rho}$ on $L^2(Q)$, any vector $\phi\in L ^2(Q)$, and any grid scheme $\{B_j^{(n)}\}_{j=1}^{N_n}$ for $Q$,} \begin{equation} \lim_{n\rightarrow\infty}\Big\langle \phi\Big|\sum_{j=1} ^{N_n}\hat{P}_j^{(n)}\hat{\rho} \hat{P}_j^{(n)}\phi\Big\rangle=0. \end{equation} {\em The same statement holds if we replace $Q$ by $\mathbb{R}^d$ (consequently setting $N_n=+\infty$).} \begin{proof} We give the proof for $Q$ and $\mathbb{R}^d$ together. The only difference will be whether $N_n$ is finite or infinite for every $n\in\mathbb{N}$. By Theorems 1.1 and 1.2 in Ref.~\cite{simon.trace.ideals}, because $\hat\rho$ is a self-adjoint and trace-class (hence, compact) operator, there exists an ONB of eigenvectors $\{\psi_\ell\}_{\ell\in\mathbb{N}}\subset L^2(Q)$ (resp. in $L^2(\mathbb{R}^d)$) with associated eigenvalues $p_\ell\geq 0$, such that, \begin{equation}\label{tal} \hat{\rho}=\sum_{\ell=1}^\infty p_\ell |\psi_\ell\rangle \langle \psi_\ell|, \end{equation} where the series is taken in the operator norm. In particular, by definition, the trace-norm of $\hat{\rho}$, $\norm{\hat\rho}_{tr}$, equals $\sum_{\ell=1}^\infty p_\ell$. Using the sequential continuity of the inner product to get Eqs.~\eqref{cual} and \eqref{pascual}, \begin{align} \label{cual} \Big\langle \phi\Big|\sum_{j=1} ^{N_n}\hat{P}_j^{(n)}\hat{\rho} \hat{P}_j^{(n)}\phi\Big\rangle&=\sum_{j=1}^{N_n}\Big\langle \phi\Big|\hat{P}_j^{(n)}\hat{\rho} \hat{P}_j^{(n)}\phi\Big\rangle\\ & \stackrel{\eqref{tal}}{=}\sum_{j=1} ^{N_n} \sum_{\ell=1}^\infty p_\ell \Big \langle \phi\:\Big|\: \hat{P}_j^{(n)}\psi_\ell\Big\rangle\Big\langle \psi_\ell \:\Big|\: \hat{P}_j^{(n)} \phi\Big\rangle\\ \label{pascual} & =\sum_{j=1} ^{N_n} \sum_{\ell=1}^\infty p_\ell\ \Big|\langle \phi| \hat{P}_j^{(n)}\psi_\ell\rangle\Big|^2\\ \intertext{[considering the series as iterated integrals in the counting measure $d\nu$ of $\mathbb{N}$, since the integrand is non-negative, we can apply the Tonelli theorem (2.37 in Ref.~\cite{folland}) to switch the order of summation]} & = \sum_{\ell=1}^\infty\sum_{j=1} ^{N_n} p_\ell\Big|\langle \phi| \hat{P}_j^{(n)}\psi_\ell\rangle\Big|^2\\ \intertext{[we define $h_n(\ell):=\sum_{j=1} ^{N_n} p_\ell\Big|\langle \phi| \hat{P}_j^{(n)}\psi_\ell\rangle\Big|^2$ and write the resulting series as an integral]} &\label{azkena} =\int_{\ell\in\mathbb{N}}h_n(\ell)d\nu. \end{align} Finally, note that: \begin{enumerate} \item[(i)] Point-wise, for each fixed $\ell\in\mathbb{N}$,\vspace{-0.3cm} \begin{equation} \lim_{n\rightarrow\infty}h_n(\ell)\ \stackrel{\text{(by def.)}}{=}\ p_\ell\lim_{n\rightarrow\infty} \sum_{j=1}^{N_n} \Big|\langle \phi| \hat{P}_j^{(n)}\psi_\ell\rangle\Big|^2\ \stackrel{\text{(Lemmas \ref{main.result.in.Q.for.pure.states} \& \ref{general.result.in.Rd.for.pure})}}{=}\ 0.\vspace{-0.3cm} \end{equation} \item[(ii)] For each $\ell\in \mathbb{N}$ and independently of $n\in\mathbb{N}$,\vspace{-0.2cm} \begin{equation} |h_n(\ell)|\stackrel{\text{(Cauchy-Schwarz)}}\leq p_\ell \norm{\phi}^2\sum_{j=1}^{N_n} \norm{\hat{P}_j^{(n)}\psi_\ell}^2\ \stackrel{\eqref{claim}}{=}\ p_\ell\norm{\phi}^2\norm{\psi_\ell}^2\ \stackrel{(\norm{\psi_\ell}=1)}{=}\ p_\ell\norm{\phi}^2. \end{equation} In particular, defining $g(\ell):=p_\ell\norm{\phi}^2$, \begin{equation} \norm{g}_{L^1(\mathbb{N},d\nu)}\ =\ \norm{\phi}^2\sum_{\ell=1}^\infty |p_\ell|\ =\ \norm{\phi}^2\norm{\hat\rho}_{tr}\ <\ +\infty. \end{equation} Hence, $g$ is a dominating function in $L^1(\mathbb{N},d\nu)$ for the sequence of functions $h_n$. \end{enumerate} By (i) and (ii), we can use the dominated convergence theorem to get the last step of the following: \begin{equation} \lim_{n\rightarrow\infty} \Big\langle \phi\Big|\sum_{j=1} ^{N_n}\hat{P}_j^{(n)}\hat{\rho} \hat{P}_j^{(n)}\phi\Big\rangle\ \stackrel{\eqref{azkena}}{=}\ \lim_{n\rightarrow\infty}\int_{\ell\in\mathbb{N}}h_n(\ell)d\nu\ =\ 0.\vspace{-0.5cm} \end{equation} \end{proof} \printbibliography \ \end{document}